Key Concepts and Formulas

• Separable Differential Equations: A differential equation of the form $g(y)dy = h(x)dx$ can be solved by integrating both sides: $\int g(y)dy = \int h(x)dx$.
• Integration Techniques: Basic integration rules are needed, particularly $\int \frac{1}{y} dy = \ln|y| + C$ and partial fraction decomposition.
• Limits: Understanding how to evaluate limits, especially as $x$ approaches 0.

Step-by-Step Solution

Step 1: Separate the variables

The given differential equation is $y(x+1)dx - x^2dy = 0$. We want to rewrite this in the form $g(y)dy = h(x)dx$. $y(x+1)dx = x^2dy$ $\frac{dy}{y} = \frac{x+1}{x^2}dx$ This separates the variables, with $y$ on the left and $x$ on the right.

Step 2: Integrate both sides

Now we integrate both sides of the equation: $\int \frac{dy}{y} = \int \frac{x+1}{x^2}dx$ $\int \frac{dy}{y} = \int \left(\frac{x}{x^2} + \frac{1}{x^2}\right)dx$ $\int \frac{dy}{y} = \int \left(\frac{1}{x} + \frac{1}{x^2}\right)dx$ Integrating both sides, we get: $\ln|y| = \ln|x| - \frac{1}{x} + C$ where $C$ is the constant of integration.

Step 3: Apply the initial condition

We are given the initial condition $y(1) = e$. This means when $x=1$, $y=e$. Substituting these values into the equation: $\ln|e| = \ln|1| - \frac{1}{1} + C$ $1 = 0 - 1 + C$ $C = 2$

Step 4: Substitute the value of C and solve for y

Substitute $C=2$ back into the equation: $\ln|y| = \ln|x| - \frac{1}{x} + 2$ Since we are considering the limit as $x \to 0^+$, we can assume $x>0$ and $y>0$. Thus we can drop the absolute value signs. $\ln y = \ln x - \frac{1}{x} + 2$ Now, exponentiate both sides to solve for $y$: $y = e^{\ln x - \frac{1}{x} + 2}$ $y = e^{\ln x} \cdot e^{-\frac{1}{x}} \cdot e^2$ $y = x e^{-\frac{1}{x}} e^2$ $y = e^2 x e^{-\frac{1}{x}}$ So, $f(x) = e^2 x e^{-\frac{1}{x}}$.

Step 5: Evaluate the limit

We want to find $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} e^2 x e^{-\frac{1}{x}}$. We can rewrite this as: $\lim_{x \to 0^+} e^2 \frac{x}{e^{\frac{1}{x}}}$ As $x \to 0^+$, $\frac{1}{x} \to \infty$, so $e^{\frac{1}{x}} \to \infty$. Thus, we have the indeterminate form $\frac{0}{\infty}$, which equals 0. $\lim_{x \to 0^+} e^2 \frac{x}{e^{\frac{1}{x}}} = e^2 \cdot 0 = 0$

Step 6: Re-examine the initial condition

The limit obtained does not match the correct answer. Let's examine our separation of variables and integration. The steps look correct. However, the correct answer suggests that the limit should be $e^2$. We made a mistake during the separation of variables.

Let's go back to Step 1: $y(x+1)dx - x^2dy = 0$ $y(x+1)dx = x^2dy$ $\frac{dy}{y} = \frac{x+1}{x^2}dx$ This step is indeed correct.

Let's check Step 2: $\int \frac{dy}{y} = \int \frac{x+1}{x^2}dx$ $\int \frac{dy}{y} = \int (\frac{1}{x} + \frac{1}{x^2})dx$ $\ln|y| = \ln|x| - \frac{1}{x} + C$

Step 3: $\ln|e| = \ln|1| - \frac{1}{1} + C$ $1 = 0 - 1 + C$ $C = 2$

Step 4: $\ln y = \ln x - \frac{1}{x} + 2$ $y = e^{\ln x - \frac{1}{x} + 2}$ $y = e^{\ln x} e^{-\frac{1}{x}} e^2$ $y = xe^{-\frac{1}{x}}e^2$ $y = e^2xe^{-\frac{1}{x}}$

Step 5: $\lim_{x \to 0^+} e^2xe^{-\frac{1}{x}} = \lim_{x \to 0^+} \frac{e^2x}{e^{\frac{1}{x}}}$. Let $u = \frac{1}{x}$. Then as $x \to 0^+$, $u \to \infty$. Also, $x = \frac{1}{u}$. So the limit becomes: $\lim_{u \to \infty} \frac{e^2(\frac{1}{u})}{e^u} = \lim_{u \to \infty} \frac{e^2}{ue^u} = 0$

It seems there is an error in the given answer. Let's try L'Hopital's rule: $\lim_{x \to 0^+} \frac{e^2x}{e^{\frac{1}{x}}} = e^2 \lim_{x \to 0^+} \frac{1}{e^{\frac{1}{x}}(-\frac{1}{x^2})} = e^2 \lim_{x \to 0^+} \frac{-x^2}{e^{\frac{1}{x}}} = 0$

Let's use L'Hopital's rule on $\lim_{u \to \infty} \frac{e^2}{ue^u}$. Then the limit becomes $\lim_{u \to \infty} \frac{0}{e^u + ue^u} = 0$.

The correct answer given is $e^2$. Let's find where the mistake lies. The mistake is in the original differential equation. The correct answer should be 0.

$\ln y = \ln x - \frac{1}{x} + 2$ $y = e^{\ln x - \frac{1}{x} + 2} = x e^{-\frac{1}{x}} e^2$ $\lim_{x \to 0^+} x e^{-\frac{1}{x}} e^2 = e^2 \lim_{x \to 0^+} \frac{x}{e^{\frac{1}{x}}}$ Apply L'Hopital's Rule $e^2 \lim_{x \to 0^+} \frac{1}{e^{\frac{1}{x}}(-\frac{1}{x^2})} = e^2 \lim_{x \to 0^+} \frac{-x^2}{e^{\frac{1}{x}}} = 0$.

The correct answer is 0. There may be a typo in the correct answer.

Common Mistakes & Tips

• Always double-check the separation of variables and integration steps.
• Be careful with indeterminate forms when evaluating limits.
• L'Hopital's rule can be useful for evaluating limits of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$.

Summary

We solved the separable differential equation by separating the variables, integrating both sides, and applying the initial condition to find the particular solution. We then evaluated the limit as $x$ approaches $0$ from the right. The limit is 0.

Final Answer

The final answer is \boxed{0}, which corresponds to option (B).